Finding Roots

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Questions 1 - 10

1

Find the sum of the solutions to:

$\frac{-6}{x^2} + \frac{1}{x} + 1 = 0$

Explanation

Multiply both sides of the equation by , to get

This can be factored into the form

So we must solve

and

to get the solutions.

The solutions are:

and their sum is .

2

Solve the following equation by factoring.

$x=0\ or\ -2$

$x=0\ or\ 2$

$x= 0,2\ or\ -2$

$x=0,-4\ or\ 1$

Explanation

First, we can factor an term out of all of the values.

We can factor remaining polynomial by determining the terms that will multiply to +4 and add to +4.

$2*2=4\ and\ 2+2=4$

Our factors are +2 and +2.

Now we can set each factor equal to zero and solve for the root.

$x=0\ or\ (x+2)^2=0$

$x=0\ or\ -2$

3

$\textup{What are the possible values of }x \textup{ when }; ; x^{2}-2x-15=0:\textup{?}$

Explanation

$\textup{Factor to solve. The product of the constants in the two roots is -15, while }$

$\textup{their sum is -2. }::\left ( x+3\right )\left ( x-5\right )=0;;;;x=-3 \textup{ or }x=5$

4

Find the root(s) of the following quadratic polynomial.

$f(x) = x^{2} + 4x + 4$

Explanation

$f(x) = x^{2} + 4x + 4$

We set the function equal to 0 and factor the equation. By FOIL, we can confirm that is equivalent to the given function. Thus, the only zero comes from, and . Thus, is the only root.

5

Solve $x^{2}-8x+12=0$ .

Explanation

Factor the quadratic equation and set each factor equal to zero:

$x^{2}-8x+12=0$ becomes so the correct answer is .

6

Solve the quadratic equation using any method:

$x=\frac{-c \pm \sqrt{c^2-4ab}}{2b}$

$x=\frac{-2c \pm \sqrt{c^2-4ab}}{2b}$

$x=\frac{-c \pm \sqrt{4c^2-4ab}}{2b}$

$x=\frac{-c \pm \sqrt{c^2-2ab}}{2b}$

$x=\frac{-c \pm \sqrt{c^2-4ab}}{b}$

Explanation

Use the quadratic formula to solve:

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x = \frac{-(c) \pm \sqrt{(c)^2-4(a)(b)}}{2(b)}$

$x = \frac{-c \pm \sqrt{c^2-4ab}}{2b}$

7

Solve the following equation using the quadratic form:

$x-4\sqrt{x}-32=0$

Explanation

Factor and solve:

$x-4\sqrt{x}-32=0$

$(\sqrt{x}-8)(\sqrt{x}+4)=0$

$\sqrt{x}=8$

or

$\sqrt{x}=-4$

This has no solutions.

Therefore there is only one solution:

8

Solve the following equation using the quadratic form:

$x=\pm 3, \pm \sqrt{3}$

$x=\pm 2, \pm \sqrt{2}$

$x=\pm 5, \pm \sqrt{5}$

$x=\pm 6, \pm \sqrt{6}$

$x=\pm 7, \pm \sqrt{7}$

Explanation

Factor and solve:

$x= \pm3$

or

$x=\pm \sqrt{3}$

Therefore the equation has four solutions:

$x= \pm3, \pm \sqrt{3}$

9

Solve the following equation using the quadratic form:

$x^{\frac{2}{3}}-9x^{\frac{1}{3}}+20=0$

Explanation

Factor and solve:

$x^{\frac{2}{3}}-9x^{\frac{1}{3}}+20=0$

$(x^{\frac{1}{3}})^2-9x^{\frac{1}{3}}+20=0$

$(x^\frac{1}{3}-5)(x^\frac{1}{3}-4)=0$

$x^\frac{1}{3}=5$

or

$x^\frac{1}{3}=4$

Therefore the equation has two solutions.

10

Find the zeros.

Explanation

Set both expressions equal to . The first factor yields . The second factor gives you .

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